Dinamica de Sistemas

INSTITUTO TECNOLOGICO DE SALTILLO

DINAMICA DE SISTEMASEJRCICIOS

DOC.JOSE G.ALVAREZ LEAL

CORREA VILLA RICARDO HELAMÁN09050317

MECATRONICA

PROBLEMA 1

M= 200B= 25K= 25

Substituyendo queda:

𝐺𝑠=𝑌 (𝑠)𝑈 (𝑠)

= 1200𝑠2+25 𝑠+25

Polos, ceros y ganancia utilizando matlaba:»syms s» num=[1];» den=[200 25 25];» [ceros,polos,gan] = tf2zp (num,den)

ceros =

Empty matrix: 0-by-1

polos =

-0.0625 + 0.3480i -0.0625 - 0.3480i

gan =

0.0050

Para graficar polos y ceros en matlab :

>> pzmap(num,den)

Grafica de polos y ceros

Ecuaciones de estado con matlab>> [A,B,C,D]=tf2ss(num,den)

A =

-0.1250 -0.1250 1.0000 0

B =

1 0

C =

0 0.0050

D =

0

Respuesta del sistema a una señal de impulso

» num=[1];» den=[200 25 25];» sys=tf(num,den)

Transfer function: 1-------------------200 s^2 + 25 s + 25

» impulse(sys)

Respuesta del sistema a una señal de escalón

» num=[1];» den=[200 25 25];» sys=tf(num,den)

Transfer function: 1-------------------200 s^2 + 25 s + 25

» step(sys)

Valores y vectores característicos del sistema

» eig(A)

ans =

-0.0625 + 0.3480i -0.0625 - 0.3480i

» [V,D]=eig(A)V =

-0.0589 + 0.3281i -0.0589 - 0.3281i 0.9428 0.9428

D =

-0.0625 + 0.3480i 0 0 -0.0625 - 0.3480i

Análisis de bode del sistema

» num=[1];» den=[200 25 25];» bode(num,den)

Problema 2

M= 200K1=40K2= 50



= 1200𝑠2+(40+50)



>> pzmap(num,den)

ceros =


polos =

0 + 0.6708i 0 - 0.6708i

gan =

0.0050



A =

0 -0.4500 1.0000 0

B =

1 0

C =

0 0.0050

D =

0


» num=[1];» den=[200 0 90];>> sys=tf(num,den) Transfer function: 1------------200 s^2 + 90 >> impulse(sys)


» num=[1];» den=[200 0 90];>> sys=tf(num,den) Transfer function: 1------------200 s^2 + 90 >> step(sys)


>> eig(A)

ans =

0 + 0.6708i 0 - 0.6708i

>> [V,D]=eig(A)

V =

0 + 0.5571i 0 - 0.5571i 0.8305 0.8305

D =

0 + 0.6708i 0 0 0 - 0.6708i



Problema 3

I= 1 Por ser el momento de inerciaC=100K= 50


𝐺𝑠=ɵ (𝑠)𝑇 (𝑠 )

= 1𝑠2+100𝑠+50


ceros =


polos =

-99.4975 -0.5025

gan =

1


>> pzmap(num,den)



A =

-100 -50 1 0

B =

1 0

C =

0 1

D =

0


>> sys=tf(num,den) Transfer function: 1----------------s^2 + 100 s + 50 >> impulse(sys)

>> sys=tf(num,den) Transfer function: 1----------------s^2 + 100 s + 50 >> step(sys)



>> eig(A)

ans =

-99.4975 -0.5025

>> [V,D]=eig(A)

V =

-0.9999 0.4490 0.0100 -0.8935

D =

-99.4975 0 0 -0.5025


>> bode(num,den)

Problema 4

• M= 200• B= 25• K= 25


𝐺𝑠=𝑋 0(𝑠)𝑋 1(𝑠)

= 25𝑠+25200𝑠2+25 𝑠+25

Polos, ceros y ganancia utilizando matlaba:

>> syms s>> num=[25 25];>> den=[200 25 25];>> [ceros,polos,gan] = tf2zp (num,den)

ceros =

-1

polos =

-0.0625 + 0.3480i -0.0625 - 0.3480i

gan =

0.1250


>> pzmap(num,den)



A =

-0.1250 -0.1250 1.0000 0

B =

1 0

C =

0.1250 0.1250

D =

0


>> sys=tf(num,den) Transfer function: 25 s + 25-------------------200 s^2 + 25 s + 25 >> impulse(sys)

>> sys=tf(num,den) Transfer function: 25 s + 25-------------------200 s^2 + 25 s + 25 >> step(sys)

Respuesta del sistema a una señal de escalon


>> eig(A)

ans =

-0.0625 + 0.3480i -0.0625 - 0.3480i

>> [V,D]=eig(A)

V =

-0.0589 + 0.3281i -0.0589 - 0.3281i 0.9428 0.9428

D =

-0.0625 + 0.3480i 0 0 -0.0625 - 0.3480i


>> bode(num,den)

Problema 5

M=1 kgC=4K=68

La ecuación que gobierna nuestro sistema es:

Aplicando Laplace queda:

Substituyendo valores queda:

Polos, ceros y ganancia utilizando matlaba:» num=[1];» den=[1 4 68];» [ceros,polos,gan]=tf2zp(num,den)

ceros =


polos =

-2.0000 + 8.0000i -2.0000 - 8.0000i

gan =

1


» pzmap(num,den)

Ecuaciones de estado con matlab» [A,B,C,D]=tf2ss(num,den)A =

-4 -68 1 0

B =

1 0

C =

0 1

D =

0


» num=[1];» den=[1 4 68];» sys=tf(num,den) Transfer function: 1--------------s^2 + 4 s + 68 » impulse(sys)


» num=[1];» den=[1 4 68];» sys=tf(num,den) Transfer function: 1--------------s^2 + 4 s + 68 » step(sys)


» eig(A)

ans =

-2.0000 + 8.0000i -2.0000 - 8.0000i

» [V,D]=eig(A)

V =

-0.9631 - 0.2408i -0.9631 + 0.2408i 0 + 0.1204i 0 - 0.1204i

D =

-2.0000 + 8.0000i 0 0 -2.0000 - 8.0000i



Problema 6

• R= 100• L= 25• C= 0.5


𝐺𝑠=𝑉𝑐(𝑠)𝑉 (𝑠)

= 1(0.5 ) (25 )𝑠2+ (1 00 ) ( 0.5 )𝑠+1


>> syms s>> num=[1];>> den=[12.5 50 1];>> [ceros,polos,gan] = tf2zp (num,den)

ceros =


polos =

-3.9799 -0.0201

gan =

0.0800


>> pzmap(num,den)



A =

-4.0000 -0.0800 1.0000 0

B =

1 0

C =

0 0.0800

D =

0


>> sys=tf(num,den) Transfer function: 1-------------------12.5 s^2 + 50 s + 1 >> impulse(sys)


>> sys=tf(num,den) Transfer function: 1-------------------12.5 s^2 + 50 s + 1 >> step(sys)


>> eig(A)

ans =

-3.9799 -0.0201

>> [V,D]=eig(A)

V =

-0.9699 0.0201 0.2437 -0.9998

D =

-3.9799 0 0 -0.0201


>> bode(num,den)

Problema 7

C1=0.005 faradsC2=0.002 faradsR1=100 hR2=500 h

Las ecuaciones que gobierna nuestro sistema son:



Polos, ceros y ganancia utilizando matlaba:» num=[1];» den=[0.5 1.75 1];» [ceros,polos,gan]=tf2zp(num,den)

ceros =


polos =

-2.7808 -0.7192

gan =

2


» pzmap(num,den)


-3.5000 -2.0000 1.0000 0

B =

1 0

C =

0 2

D =

0


» num=[1];» den=[0.5 1.75 1];» sys=tf(num,den) Transfer function: 1--------------------0.5 s^2 + 1.75 s + 1 » impulse(sys)


» num=[1];» den=[0.5 1.75 1];» sys=tf(num,den) Transfer function:

1--------------------0.5 s^2 + 1.75 s + 1 » step(sys)


» eig(A)

ans =

-2.7808 -0.7192

» [V,D]=eig(A)

V =

-0.9410 0.5839 0.3384 -0.8118

D =

-2.7808 0 0 -0.7192


» num=[1];» den=[0.5 1.75 1];» bode(num,den)

Problema 8• MODELACIÓN MATEMÁTICA

Suspensión de un automóvil

f(t)

z(t)

kb

m

Fuerza de entrada

Desplazamiento, salida del sistema

• M= 200• K=25• B=25

2

2 )()()()(

dt

tzdm

dt

tdzbtkztf

maF

kbsmssF

sZ

kbsmssZsF

sZmssbsZskZsF

dt

tzdm

dt

tdzbtkztf

2

2

2

2

2

1

)(

)(

)()(

)()()()(

cero) a igual iniciales scondicione ndo(considera

términocada a Laplace de ada transformla Aplicando

)()()()(


= 1200𝑠2+25 𝑠+25



>> syms s>> num=[1];>> den=[200 25 25];>> [ceros,polos,gan] = tf2zp (num,den)

ceros =


polos =

-0.0625 + 0.3480i -0.0625 - 0.3480i

gan =

0.0050


>> pzmap(num,den)



A =

-0.1250 -0.1250 1.0000 0

B =

1 0

C =

0 0.0050

D =

0


>> sys=tf(num,den) Transfer function: 1-------------------200 s^2 + 25 s + 25 >> impulse(sys)


>> sys=tf(num,den) Transfer function: 1-------------------200 s^2 + 25 s + 25 >> step(sys)


>> eig(A)

ans =

-0.0625 + 0.3480i -0.0625 - 0.3480i

>> [V,D]=eig(A)

V =

-0.0589 + 0.3281i -0.0589 - 0.3281i 0.9428 0.9428

D =

-0.0625 + 0.3480i 0 0 -0.0625 - 0.3480i


>> bode(num,den)

Problema 9

Nivel en un tanque

qo(t)

Flujo de salida

R

(resistencia de la válvula)

h(t)

qi(t)

Flujo de entrada

Flujo que entra – Flujo que sale = Acumulamiento

A

(área del tanque)

• R= 50• A=150

dt

tdhAth

Rtq

tq

thR

dt

tdhAtqtq

i

o

oi

)()(

1)(

)(

)(

)()()(

111

)(

)(

)1

)(()(

)()(1

)(

Laplace de ada transformla Aplicando

)()(

1)(

ARs

R

RAssQ

sHR

AssHsQi

sAsHsHR

sQi

dt

tdhAth

Rtq

i

i


𝐺𝑠=𝐻 (𝑠)𝑄𝑖(𝑠)

= 507500 𝑠❑+1


>> syms s>> num=[50];>> den=[7500 1];>> [ceros,polos,gan] = tf2zp (num,den)

ceros =


polos =

-1.3333e-004

gan =

0.0067


>> pzmap(num,den)



A =

-1.3333e-004

B =

1

C =

0.0067

D =

0


>> sys=tf(num,den) Transfer function: 50----------7500 s + 1 >> impulse(sys)


>> sys=tf(num,den) Transfer function: 50----------7500 s + 1 >> step(sys)


>> eig(A)

ans =

-1.3333e-004

>> [V,D]=eig(A)

V =

1

D =

-1.3333e-004


>> bode(num,den)

Problema 10

• K1=100• M1= 50• B=25• K12=75• M2=50

>> syms s>> num=[50 25 75];>> den=[50 2500 10000 5000 3900];>> [ceros,polos,gan] = tf2zp (num,den)



𝐺𝑠=𝐻 (𝑠)𝑄𝑖(𝑠)

= 50+25+7550 𝑠4 +2500𝑠3❑+10000 𝑠2+5000 𝑠❑+3900


>> pzmap(num,den)

ceros =

-0.2500 + 1.1990i -0.2500 - 1.1990i

polos =

-45.6677 -3.8921 -0.2201 + 0.6248i -0.2201 - 0.6248i

gan =

1



A =

-50 -200 -100 -78 1 0 0 0 0 1 0 0 0 0 1 0

B =

1 0 0 0

C =

0 1.0000 0.5000 1.5000

D =

0


>> sys=tf(num,den) Transfer function: 50 s^2 + 25 s + 75---------------------------------------------50 s^4 + 2500 s^3 + 10000 s^2 + 5000 s + 3900 >> impulse(sys)

>> sys=tf(num,den) Transfer function: 50 s^2 + 25 s + 75---------------------------------------------50 s^4 + 2500 s^3 + 10000 s^2 + 5000 s + 3900 >> step(sys)


Valores y vectores característicos del sistema>> eig(A)

ans =

-45.6677 -3.8921 -0.2201 + 0.6248i -0.2201 - 0.6248i

>> [V,D]=eig(A)

V =

0.9998 0.9664 0.1887 - 0.1169i 0.1887 + 0.1169i -0.0219 -0.2483 -0.2610 - 0.2100i -0.2610 + 0.2100i 0.0005 0.0638 -0.1680 + 0.4770i -0.1680 - 0.4770i -0.0000 -0.0164 0.7634 0.7634

D =

-45.6677 0 0 0 0 -3.8921 0 0 0 0 -0.2201 + 0.6248i 0 0 0 0 -0.2201 - 0.6248i


>> bode(num,den)

Problema 11

b= 10M= 100 K= 10



Polos, ceros y ganancia utilizando matlaba:» num=[10 10];» den=[100 10 10];» [ceros,polos,gan] = tf2zp (num,den)


ceros =

-1

polos =

-0.0500 + 0.3122i -0.0500 - 0.3122i

gan =

0.1000Para graficar polos y ceros en matlab :

>> pzmap(num,den)


Ecuaciones de estado con matlab» [A,B,C,D]=tf2ss(num,den)

A =

-0.1000 -0.1000 1.0000 0

B =

1 0

C =

0.1000 0.1000

D =

0


» num=[10 10];» den=[100 10 10];» sys=tf(num,den) Transfer function:

10 s + 10-------------------100 s^2 + 10 s + 10

» impulse(sys)


» num=[0 10 10];» den=[100 10 10];» sys=tf(num,den) Transfer function:

10 s + 10-------------------100 s^2 + 10 s + 10

» step(sys)


» eig(A)

ans =

-0.0500 + 0.3122i -0.0500 - 0.3122i

» [V,D]=eig(A)

V =

-0.2977 - 0.0477i -0.2977 + 0.0477i 0 + 0.9535i 0 - 0.9535i

D =

-0.0500 + 0.3122i 0 0 -0.0500 - 0.3122i


» num=[10 10];» den=[100 10 10];» bode(num,den)

Problema 12

Esta es la ecuación que representa nuestro el sistema:

Aplicando Laplace:

Queda un sistema de ecuaciones algebraico donde se puede eliminarla variable X1. De esta forma la función de transferencia del sistema es:

Evidentemente, también se podría haber tomado X1 como salida del sistema y eliminar variable X2. En este caso, la función de transferencia que se obtendría es:

Desarrollando el denominadora las funciones de transferencia quedarían La siguiente manera:

Substituyendo valores las funciones de transferencia quedarían La siguiente manera:

Polos, ceros y ganancia utilizando matlaba:» num1=[100 200];» den1=[200 300 390 -100 1000];» [ceros,polos,gan]=tf2zp(num1,den1)

ceros =-2

polos = -1.3253 + 1.3644i -1.3253 - 1.3644i 0.5753 + 1.0252i 0.5753 - 1.0252i

gan =0.5000


Ecuaciones de estado con matlab» [A,B,C,D]=tf2ss(num1,den1)

A = -1.5000 -1.9500 0.5000 -5.0000 1.0000 0 0 0 0 1.0000 0 0 0 0 1.0000 0B =

1 0 0 0C =

0 0 0.5000 1.0000D = 0


» num1=[100 200];» den1=[200 300 390 -100 1000];» sys1=tf(num1,den1) Transfer function: 100 s + 200------------------------------------------200 s^4 + 300 s^3 + 390 s^2 - 100 s + 1000

» impulse(sys1)


» num1=[100 200];» den1=[200 300 390 -100 1000];» sys1=tf(num1,den1) Transfer function: 100 s + 200------------------------------------------200 s^4 + 300 s^3 + 390 s^2 - 100 s + 1000

step(sys1)


» eig(A)ans = -1.3253 + 1.3644i -1.3253 - 1.3644i 0.5753 + 1.0252i 0.5753 - 1.0252i» [V,D]=eig(A)V = 0.7445 + 0.4165i 0.7445 - 0.4165i -0.1324 + 0.6027i -0.1324 - 0.6027i -0.1157 - 0.4334i -0.1157 + 0.4334i 0.3920 + 0.3491i 0.3920 - 0.3491i -0.1211 + 0.2024i -0.1211 - 0.2024i 0.4222 - 0.1454i 0.4222 + 0.1454i 0.1207 - 0.0285i 0.1207 + 0.0285i 0.0679 - 0.3737i 0.0679 + 0.3737i

D =

-1.3253 + 1.3644i 0 0 0 0 -1.3253 - 1.3644i 0 0 0 0 0.5753 + 1.0252i 0 0 0 0 0.5753 - 1.0252i


» num1=[100 200];» den1=[200 300 390 -100 1000];» bode(num1,den1)

Problema 13

K1 = 100K2 = 150B2 = 200M= 1000



Polos, ceros y ganancia utilizando matlaba:» num=[200 150 100];» den=[1000 200 150 100];» [ceros,polos,gan]=tf2zp(num,den)ceros =

-2 -1polos =

-7.7417 -0.2583

gan =

1


» pzmap(num,den)


-8 -2 1 0

B =

1 0

C =

-5 0

D =

1


» num=[200 150 100];» den=[1000 200 150 100];» sys=tf(num,den) Transfer function:

200s^2 + 150 s + 100------------------- 1000s^3 + 200 s^2 + 150s+100

» impulse(sys)


» num=[200 150 100];» den=[1000 200 150 100];» sys=tf(num,den) Transfer function:

0.5 s^2 + 1.5 s + 1------------------- 0.5 s^2 + 4 s + 1

» step(sys)


» eig(A)

ans =

-7.7417 -0.2583

» [V,D]=eig(A)

V =

-0.9918 0.2501 0.1281 -0.9682

D =

-7.7417 0 0 -0.2583



Problema 14

C1=0.005 faradsC2=0.002 faradsR1=100 hR2=500 h



Polos, ceros y ganancia utilizando matlaba:» num=[0.5 1.5 1];» den=[0.5 4 1];» [ceros,polos,gan]=tf2zp(num,den)ceros =

-2 -1polos =

-7.7417 -0.2583

gan =

1


» pzmap(num,den)


-8 -2 1 0

B =

1 0

C =

-5 0

D =

1


» num=[0.5 1.5 1];» den=[0.5 4 1];» sys=tf(num,den) Transfer function:

0.5 s^2 + 1.5 s + 1------------------- 0.5 s^2 + 4 s + 1

» impulse(sys)


» num=[0.5 1.5 1];» den=[0.5 4 1];» sys=tf(num,den) Transfer function:

0.5 s^2 + 1.5 s + 1------------------- 0.5 s^2 + 4 s + 1

» step(sys)


» eig(A)

ans =

-7.7417 -0.2583

» [V,D]=eig(A)

V =

-0.9918 0.2501 0.1281 -0.9682

D =

-7.7417 0 0 -0.2583



Problema 15

• R= 100• L= 25• C= 0.5


𝐺𝑠=𝑉𝑐(𝑠)𝑉 (𝑠)

= 1

(100 ) (25 ) 𝑠2+ 1000.5

𝑠+1




>> pzmap(num,den)

ceros =


polos =

-0.0746 -0.0054

gan =

4.0000e-004



A =

-0.0800 -0.0004 1.0000 0

B =

1 0

C =

1.0e-003 *

0 0.4000

D =

0


>> sys=tf(num,den) Transfer function: 1--------------------2500 s^2 + 200 s + 1 >> impulse(sys)


>> sys=tf(num,den) Transfer function: 1--------------------2500 s^2 + 200 s + 1 >> step(sys)


>> eig(A)

ans =

-0.0746 -0.0054

>> [V,D]=eig(A)

V =

-0.0744 0.0054 0.9972 -1.0000

D =

-0.0746 0 0 -0.0054


>> bode(num,den)

Problema 16




Polos, ceros y ganancia utilizando matlaba:» num=[0.047 0.31385 0.445];» den=[0.037 0.31385 0.445];» [ceros, polos,gan]=tf2zp(num,den)

ceros =

-4.6349 -2.0428

polos =

-6.6827 -1.7997

gan =

1.2703


» pzmap(num,den)

Ecuaciones de estado con matlab» [A,B,C,D]=tf2ss(num,den)

A = -8.4824 -12.0270 1.0000 0

B = 1 0

C =

-2.2925 -3.2505

D =

1.2703


» num=[0.047 0.31385 0.445];» den=[0.037 0.31385 0.445];» sys=tf(num,den) Transfer function:0.047 s^2 + 0.3139 s + 0.445----------------------------0.037 s^2 + 0.3139 s + 0.445 » impulse(sys)


» num=[0.047 0.31385 0.445];» den=[0.037 0.31385 0.445];» sys=tf(num,den) Transfer function:0.047 s^2 + 0.3139 s + 0.445----------------------------0.037 s^2 + 0.3139 s + 0.445 » step(sys)


» eig(A)

ans =

-6.6827 -1.7997

» [V,D]=eig(A)

V =

-0.9890 0.8741 0.1480 -0.4857

D =

-6.6827 0 0 -1.7997


» num=[0.047 0.31385 0.445];» den=[0.037 0.31385 0.445];» bode(num,den)

Problema 17

B1= 200B2= 150K1= 125K2= 200

>> syms s>> num=[200 125]

num = 200 125

>> den=[350 325]den =

350 325>> sys= tf(num,den)Transfer function:

200 s + 125-----------

350 s + 325>> roots (num)

ans = -0.6250

>> roots (den)ans =

-0.9286

>> PzMap(num,den)>> [A B C D]=tf2ss(num,den)

A = -0.9286

B = 1C =

-0.1735D =

0.5714

>> impulse(sys)

>> step(sys)

>> [frec,amort]=damp(den)

frec =

0.9286

amort =

1

>> t=[0:0.1:3];

>> [x,y]=step(num,den,t)

VALORES:

>> lambda=eig(A)

lambda =

-0.9286

>> lambda=eig(B)

lambda =

1

>> lambda=eig(C)

lambda =

-0.1735

>> lambda=eig(D)

lambda =

0.5714

VECTORES:

>> [S,D]=eig(A)

S =

1

D =

-0.9286

CANONICAS:

>> [num,den]=ss2tf(A,B,C,D)

num =

-0.9286 -1.0357

den =

1.0000 0.9286

>> [A,B,C,D]=canon(A,B,C,D,'companion')

A =

-0.9286

B =

1

C =

-0.1735

D =

-0.9286

>> [A,B,C,D]=canon(A,B,C,D,'modal')

A =

-0.9286

B =

1

C =

-0.1735

D =

-0.9286

>> margin(A,B,C,D)

>> [Gm,Pm,Wcg,Wcp]=margin(num,den)

Gm =

1.0769

Pm =

-5.1470

Wcg =

Inf

Wcp =

1.2361

>> w=0:0.1:100;

>> bode(num,den,w)

Problema 18


Aplicando transformada de Laplace quedan:


Polos, ceros y ganancia utilizando matlaba:» num=[1];» den=[0.4 0.2 1];» [ceros,polos,gan]=tf2zp(num,den)ceros =


polos =

-0.2500 + 1.5612i -0.2500 - 1.5612i

gan =

2.5000


» pzmap(num,den)


-0.5000 -2.5000 1.0000 0

B =

1 0

C =

0 2.5000

D =

0


» pzmap(num,den)» num=[1];» den=[0.4 0.2 1];» sys=tf(num,den) Transfer function:

1-------------------0.4 s^2 + 0.2 s + 1

» impulse(sys)


» pzmap(num,den)» num=[1];» den=[0.4 0.2 1];» sys=tf(num,den) Transfer function:

1-------------------0.4 s^2 + 0.2 s +

» step(sys)


» eig(A)

ans =

-0.2500 + 1.5612i -0.2500 - 1.5612i

» [V,D]=eig(A)

V =

-0.8345 - 0.1336i -0.8345 + 0.1336i 0 + 0.5345i 0 - 0.5345i

D =

-0.2500 + 1.5612i 0 0 -0.2500 - 1.5612i



Problema 19




I1000.5




>> pzmap(num,den)

ceros =


polos =

-2.5000 +13.9194i -2.5000 -13.9194i

gan =

1



A =

-5 -200 1 0

B =

1 0

C =

0 1

D =

0


>> sys=tf(num,den) Transfer function: 1---------------s^2 + 5 s + 200 >> impulse(sys)


>> sys=tf(num,den) Transfer function: 1---------------s^2 + 5 s + 200 >> step(sys)


>> eig(A)

ans =

-2.5000 +13.9194i -2.5000 -13.9194i

>> [V,D]=eig(A)

V =

-0.9975 -0.9975 0.0125 + 0.0694i 0.0125 - 0.0694i

D =

-2.5000 +13.9194i 0 0 -2.5000 -13.9194i


>> bode(num,den)

Problema 20

• M= 200• K= 25

fk= kxfm1=mx"δ(t)=1mx" + kx = δ(t)x"+(k/m)x=1/m£{x"}+(k/m)£{x}=1/ms2X(s)+(K/M)X(s)= 1/MX(s) [s2+(K/M)]=1/M




X(s)


>> syms s>> num=[0.005];>> den=[1 0 8];>> [ceros,polos,gan] = tf2zp (num,den)

ceros =


polos =

0 + 2.8284i 0 - 2.8284i

gan =

0.0050


>> pzmap(num,den)



A =

0 -8 1 0

B =

1 0

C =

0 0.0050

D =

0


>> sys=tf(num,den) Transfer function: 0.005-------s^2 + 8 >> impulse(sys)


>> sys=tf(num,den) Transfer function: 0.005-------s^2 + 8 >> step(sys)


>> eig(A)

ans =

0 + 2.8284i 0 - 2.8284i

>> [V,D]=eig(A)

V =

0.9428 0.9428 0 - 0.3333i 0 + 0.3333i

D =

0 + 2.8284i 0 0 0 - 2.8284i


>> bode(num,den)

Date post:	26-Jul-2015
Category:	Documents
Upload:	eduardo-lopez-suarez
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